This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CHEM 1314 3;3O pm Theory
Exam I John I. Gelder September 11, 2002 Page 2 SCORES
(1) Page 2 (30) Name TA's Name Lab Section INSTRUCTIONS: . This examination consists of a total of 6 different pages. The last page include a periodic table and some useful
equations. All work should be done in this booklet. . PRINT your name, TA's name and your lab section number m in the space at the top of this sheet. 2Q
NOT SEPARATE THESE PAGES. . Answer all questions that you can and whenever called for show your work clearly. Your method of solving
problems should pattern the approach used in lecture. . No credit will be awarded if your work is not shown in problems 2, 4 and 7a. . Point values are shown next to the problem number. . Budget your time for each of the questions. Some problems may have a low point value yet be very
challenging. If you do not recognize the solution to a
question quickly, skip it, and return to the question after
completing the easier problems. . Look through the exam before beginning; plan your work; then begin. {relax and do well.
Page 3 Page 4 Page 5 TOTAL
(24) (29) (16) (100) CHEM 1314 EXAMI (6) 3. (18) 4. a) b) d) PAGE 3 Diagram the following system as viewed at the atomic level in the space provided. Be sure to clearly label
each of the substances in your diagram. carbon monoxide hydro gen Grading: 2 points each for the correct
diatomic for hydrogen and carbon
monoxide. 1 point for gas phase and
1 point for homogeneous mixture A homogeneous
mixture of carbon
monoxide and
hydrogen at room
temperature. A particular homogeneous mixture of ethanol (C2H50) and water has a density of 0.945 g  ml‘l. This
ethanol solution is also 36.0% (by mass) ethanol. The density of pure ethanol is 0.789 g ' ml'l. A sample of this ethanol solution has a mass of 250. g, calculate the volume of the solution. . ( 1 mL ethanol solution ) .
250. g ethanol solutlon ————_———0.945 3 ethanol solution = 265 mL of ethanol solution How many grams of water and ethanol are in the sample in a)? 36.0 e eth l
g pur am} ) = 90.0 g of pure ethanol 250. g ethanol solution (100 g ethanol solution 250. grams — 90.0 grams = 160. grams of water Calculate the volumes of pure ethanol and water that must be mixed to prepare the solution in part a)? 1 mL pure ethanol
90.0 g pure ethanol mm = 114 mL of pure ethanol 1 mL pure water
160.0 g pure water W = 160 mL of pure water What is interesting about the answer in part a) and the answer(s) in part c)? To prepare the solution of ethanol and water we would mix 114 mLs ethanol with 160 mLs of
water. What is interesting is the volume of the mixture is smaller than the volume of the pure liquids added together. So upon mixing there is a contraction of the total volume. Why this
happens is discussed in CHEM 1515. Grading: Parts 3, b and c are worth 5 points each. Pretty much R/W. 1 point for the correct units in part a). Part (1) is worth 3 points. It was relatively easy to get 1 point (assuming parts
a and c were correct), a little harder to get two points and 3 points were pretty difﬁcult. CHEM 1314 EXAM I PAGE 5 (6) 8. Complete each calculation and report the answer to the correct number of signiﬁcant ﬁgures. (3.05 x1020)_ 3 66 ms
a) 8.340x1024 ‘ ‘ X 1 1
b) + (m) = 0.1538 + 0.000310% = 0.016 , 2.001 M
c) (8.34  0.450)— 0.83 = sass—21111 = 1.3 grading: R/W 2 points each
(10) 9. Complete the following table W" Fula Vim WW" 7'" " 7' ms 0 ’ umber of atos, '7
g molecules, or formula units
Mass m sample (gms) sample (mol)
1.2mm“. 7,, 7 V1,,” X111 1,, . 1 7X fu
2 3
P205: 284 g (%%1) = 2.00 mol 2.00 mol (W) = 1.20 x 1024 f.u.
X 5 32 1022  —°
. . X atoms 6.02 X 1023 atoms — 8.84 X 10 ~ mol
8.478 g g . = 95.94 m X 15 Mo (molybdenum)
. 151 g
NaBr03: 3.21 X 103 mol (1 mol) = 0.484 g (6.02 x 102 3 f.u.
3.21 X 10—3 mol T = 1.93 X 102 1 f.u. What is the symbol for the unknoan element, X? M0 CHEM 1314 EXAM l I PAGE 4 (6) 5. Predict a reasonable formula for the compound formed from each of the following combinations of
elements or polyatomic ions. a) sodium and sulfate N azSO4
b) iron and bromide FeBrz or FeBr3
0) carbon and ﬂuorine CF4 Grading: W 2 points each (11) 6. Complete the following table;
Ionic r e Name of the compound Formula of t 7777 7 r _ , ,, Within,
, in, ,, e ,, ovalent Grading: R/W 1 point each. Spelling and symbols had to be perfect for the point. (12) 7. The relative weighted average atomic mass of gallium is 69.723 11. Gallium has two naturally occurring
isotopes. One of those isotopes has a mass of 70.9247 11 and a percent abundance of 39.89%. a) Calculate the mass of the other isotope of gallium. The relative average atomic mass of gallium from the periodic table is 69.723 11. The
fractional abundance of the unknown Ga isotope must be 1  0.3989 = 0.6011, because only
two isotopes of Ga exist. Average atomic mass = 2(massisot0pe  fractional abundanceisotope) 69.723 11 = (70.9247 u)(0.3989) + x( 0.6011)
69.723 u = 23.292 + x( 0.6011) u = x( 0.6011) x = 68.93 11 b) How many neutrons, protons and electrons are in each of the isotopes of gallium? “
6 9 31 38 31
——— CHEM 1314 EXAM I (12) 1. a)
b) C)
d) (18) 2. b) PAGE 2
Write the chemical formula(s) of the product(s) and balance all of the following reactions. Identify all
products phases as either (g)as, (Diquid, (s)olid or (aq)ueous 3Mg(s) + N2(g) —> Mg3N2(s) 58(3) + 8020.9) —> 850269) 2C6H14(l) + 1902(g) —> 12C02(g) + 14H20(g)
2K(s) + C12(g) —> 2KC1(s) Grading: 3 points for the correct products, balance and phases. If the products are ﬂ correct,
—1 for balance and —1 for phases Perform the following conversions. Be sure your answer has the correct number of significant figures. 400. cm3 to liters 400 £1113 = 0 400 L
1000 mL ' 1 c1113 the volume of a ball is 2.00 ft3. Convert to km3. (use at least 3 conversions factors.) 12 in) 3 (2.54 cm) 3 ( 3 3
200 ft3 (item 1 inch 100 cm) (1000 In) = 565" 10‘ 11 km What is 10.0 Kelvin on the Fahrenheit scale? K = °C + 273.15
°C =K —27.15 = 10.0 — 273.15 = 263.2 °C 9 9
0F = 3pc) + 32 = = §(263.2) + 32 = 441.8 Grading: each part is worth 6 points. Deduct 1 point for sig fig, 1 point for units, 1 point for
math error. Deduct 3 points for a ‘simple’ conversion error. If more than 1 conversion error
deduct all points. ...
View
Full Document
 Fall '07
 Gelder

Click to edit the document details